2y^2+3y-0.5=0

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Solution for 2y^2+3y-0.5=0 equation: 



2y^2+3y-0.5=0

a = 2; b = 3; c = -0.5;
Δ = b2-4ac
Δ = 32-4·2·(-0.5)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{13}}{2*2}=\frac{-3-\sqrt{13}}{4}
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{13}}{2*2}=\frac{-3+\sqrt{13}}{4}
$

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